In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$

Solution: Since a $13-14-15$ triangle is a $5-12-13$ triangle and a $9-12-15$ triangle "glued" together on the $12$ side, $[ABC]=\frac{1}{2}\cdot12\cdot14=84$.
There are six points of intersection between $\Delta ABC$ and $\Delta A'B'C'$. Connect each of these points to $G$.
[asy] size(8cm); pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C_1,C_2; B=(0,0); A=(5,12); C=(14,0); E=(12.6667,8); D=(7.6667,-4); F=(-1.3333,8); G=(6.3333,4); B_1=(4.6667,0); B_2=(1.6667,4); A_1=(3.3333,8); A_2=(8,8); C_1=(11,4); C_2=(9.3333,0); dot(A); dot(B); dot(C); dot(G); dot(D); dot(E); dot(F); dot(A_1); dot(B_1); dot(C_1); dot(A_2); dot(B_2); dot(C_2); draw(B--A--C--cycle); draw(E--D--F--cycle); draw(B_1--A_2); draw(A_1--C_2); draw(C_1--B_2); label("$B$",B,WSW); label("$A$",A,N); label("$C$",C,ESE); label("$G$",G,S); label("$B'$",E,ENE); label("$A'$",D,S); label("$C'$",F,WNW); [/asy]
There are $12$ smaller congruent triangles which make up the desired area. Also, $\Delta ABC$ is made up of $9$ of such triangles. Therefore, $\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}$.